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A cylindrical pitcher with a radius 5 is filled with beer to a height

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A cylindrical pitcher with a radius 5 is filled with beer to a height  [#permalink]

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06 Sep 2018, 00:57
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Difficulty:

25% (medium)

Question Stats:

85% (00:50) correct 15% (01:40) wrong based on 26 sessions

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A cylindrical pitcher with a radius 5 is filled with beer to a height of 9 inches. If the beer is poured from the first pitcher to a second pitcher with no spillage and fills the second pitcher to a height of 4 inches, what is the radius of the second pitcher?

A. 15 inches
B. 12.5 inches
C. 9 inches
D. 8 inches
E. 7.5 inches

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Re: A cylindrical pitcher with a radius 5 is filled with beer to a height  [#permalink]

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06 Sep 2018, 03:15
Bunuel wrote:
A cylindrical pitcher with a radius 5 is filled with beer to a height of 9 inches. If the beer is poured from the first pitcher to a second pitcher with no spillage and fills the second pitcher to a height of 4 inches, what is the radius of the second pitcher?

A. 15 inches
B. 12.5 inches
C. 9 inches
D. 8 inches
E. 7.5 inches

Volume of beer in first pitcher = Volume of Beer in second pitcher

Volume of a cylinder = $$πr^2h$$

i.e. $$π*5^2*9 = π*r^2*4$$
i.e. $$r = 7.5$$

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A cylindrical pitcher with a radius 5 is filled with beer to a height  [#permalink]

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06 Sep 2018, 08:06
Bunuel wrote:
A cylindrical pitcher with a radius 5 is filled with beer to a height of 9 inches. If the beer is poured from the first pitcher to a second pitcher with no spillage and fills the second pitcher to a height of 4 inches, what is the radius of the second pitcher?

A. 15 inches
B. 12.5 inches
C. 9 inches
D. 8 inches
E. 7.5 inches

$$πR^2H = πr^2h$$

Or, $$π*5^2*9 = π*r^2*4$$

Or, $$5^2*3^2 = r^2*2^2$$

Or, $$5*3 = 2r$$

So, $$r = 7.5$$, ANswer must be (E)

PS : IMHO, the highlighted part must be with a radius 5 inches Bunuel
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A cylindrical pitcher with a radius 5 is filled with beer to a height  [#permalink]

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06 Sep 2018, 11:34
Bunuel wrote:
A cylindrical pitcher with a radius 5 is filled with beer to a height of 9 inches. If the beer is poured from the first pitcher to a second pitcher with no spillage and fills the second pitcher to a height of 4 inches, what is the radius of the second pitcher?

A. 15 inches
B. 12.5 inches
C. 9 inches
D. 8 inches
E. 7.5 inches

The volume of beer is equal in both pitchers.

Volume of [the liquid in] a cylinder, and thus in the first pitcher:
$$V =\pi r^2 h$$
$$V=\pi* 5^2*9=225\pi$$

That volume of beer is in the second pitcher. Solve for radius. Second pitcher:
$$225\pi=\pi r^2h=\pi* r^2* 4$$
$$\frac{225}{4}=r^2$$
$$56.25=r^2$$
$$\sqrt{56.25}=r$$

Between which two perfect squares does $$56.25$$ lie?

$$7^2=49 < 56.25 < 64=8^2$$

The radius is between $$7$$ and $$8$$.

(C) $$9^2=81$$ -> too great. We need 56.25
(D) $$8^2=64$$ -> too great.

By POE, (E) is the answer.
And: (E) 7.5 is between 7 and 8
$$7^2 = 49$$
$$8^2=64$$
$$49<56.25<64$$ - That works.

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A cylindrical pitcher with a radius 5 is filled with beer to a height &nbs [#permalink] 06 Sep 2018, 11:34
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